Convergence in measure of sequence of functions Hi I don't have a lot of experience in measure theory so that might be basic. If you have a sequence of functions $a_{k}(x): \Omega \rightarrow \mathbb{R}$ such that 
$$0 \leq\limsup\limits_{k \rightarrow \infty}\int_{\Omega}a_{k}(x)dx \leq 0$$ How does this imply that $a_{k}(x) \rightarrow 0$ in measure? 
Thanks.
 A: In general, the statement is not correct. Simply consider $a_k := a$ for some $a: \Omega \to \mathbb{R}$ such that $$\int_{\Omega} a(x) \, dx=0.$$
However, the statement is correct if $a_k$ is a sequence of non-negative functions as the following proof shows:
Proof: Let $(a_k)_{k \in \mathbb{N}}$ be a sequence of non-negative functions and denote by $\lambda$ the Lebesge measure. From
$$0 \leq \liminf_{k \to \infty} \int_{\Omega} a_k(x) \, \lambda(dx) \leq \limsup_{k \to \infty} \int_{\Omega} a_k(x) \, \lambda(dx) \leq 0$$
it follows that the limit $$\lim_{k \to \infty} \int_{\Omega} a_k(x) \, \lambda(dx)$$ (exists and) equals $0$. This means that $a_k$ converges in $L^1$ to $0$. On the other hand, Markov's inequality implies
$$\begin{align*} \lambda(\{x \in \Omega; |a_k(x)| \geq \varepsilon\} &= \lambda(\{x \in \Omega; a_k(x) \geq \varepsilon\}) \\ &\leq \frac{1}{\varepsilon} \int_{\Omega} a_k(x) \, dx. \end{align*}$$
Letting $k \to \infty$, we see that the left-hand side converges to $0$. Hence, $a_k \to 0$ in measure.
